Question

我有一个简单的PHP文件上传脚本，如此，

<?php

    $path = 'uploads/';
    $file_ext   = array('jpg','png','gif','bmp','JPG');
    $post_ext   = end(explode('.',$_FILES['photo']['name']));
    $photo_name = $_FILES['photo']['name'];
    $photo_type = $_FILES['photo']['type'];
    $photo_size = $_FILES['photo']['size'];
    $photo_tmp  = $_FILES['photo']['tmp_name'];
    $photo_error= $_FILES['photo']['error'];
    //move_uploaded_file($photo_tmp,"uploads/".$photo_name);

    echo $photo_tmp;

    if((($photo_type == 'image/jpeg') || ($photo_type == 'image/gif')   ||
       ($photo_type == 'image/png') || ($photo_type == 'image/pjpeg')) &&
       ($photo_size < 2000000) && in_array($post_ext,$file_ext)) {

        /* Understand in-Array !! */

        if($photo_error > 0 ){
            echo 'Error '.$photo_error;
            exit;
        }else{
            echo $photo_name.' Uploaded !';
        }
        if(file_exists($path.$photo_name)){
            echo 'There is '.$photo_name;
        }else{
            //new photo name and encryption
            $new_name = explode('.',$photo_name);
            $photo_name = 'erkan_'.md5($new_name[0]).'.'.$new_name[1];

            //move to directory
            if(move_uploaded_file($photo_tmp,$path.$photo_name)){

                return $photo_name;
            }
        }
    }

?>

表单代码：

<form action="fileupload.php" method="post" enctype="multipart/form-data">
            <input type="file" name="photo" id="fileBox">
            <button type="submit">SUBMIT</button>

使用上面的脚本我的文件保存完好，但是当我在混合中添加一点ajax时，就像这样：

$(function(){
    $('button[type="submit"]').on('click' , function(e){
        e.preventDefault();
        var formData = new FormData();
        formData.append('photo', $('input[type=file]')[0].files[0]); 
        $.ajax({
            url: 'fileupload.php',
            data: formData,
            // THIS MUST BE DONE FOR FILE UPLOADING
            contentType: false,
            processData: false,
            // ... Other options like success and etc
        });
    });
});

现在，当我上传图片时，图片未保存在我的上传文件夹中，为什么？

Answer 1

您在type: "POST"标记中遗漏了<script></script>。

$(function(){
    $('button[type="submit"]').on('click' , function(e){
        e.preventDefault();
        var formData = new FormData();
        formData.append('photo', $('input[type=file]')[0].files[0]); 
        $.ajax({
            type: "POST",  
            url: 'fileupload.php',
            data: formData,
            // THIS MUST BE DONE FOR FILE UPLOADING
            contentType: false,
            processData: false,
            // ... Other options like success and etc
        });
    });
});

Answer 2

你在ajax调用中有一些错误。

1 - 我为您的表单提供了ID

2 - 您没有从表单中获取值。

$(function(){
    $('#submitBtn').on('click' , function(e){
        e.preventDefault();
            var form = $('form#myForm');

            var formdata = false;

            if (window.FormData){
                formdata = new FormData(form[0]);
            }

            var formAction = form.attr('action');

            $.ajax({
                url: formAction,
                data : formdata ? formdata : form.serialize(),
                cache : false,
                contentType : false,
                processData : false,
                dataType: "json",
                type : 'POST',
                resetForm: true,
            })
            .done(function(data) {
               //do something with the returned data
            }
     });
 });

Answer 3

我认为您的文件名未正确设置。所以你的PHP无法获取该文件。试试这个

 formData.append('photo', $('input[type=file]')[0].files[0],$('input[type=file]')[0].files[0].name);

当我向代码添加ajax时图像不保存

3 个答案: